A battery of emf 10 V and internal resistance 3`Omega ` is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor ? What is the terminal voltage of the battery when the circuit is closed ?

Explanation

Here `epsi = 10 V,r= 3Omega` and I = 0.5 A
As `I = (epsi)/((R + r))` , hence external resistance R = `epsi/I - r = (10)/(0.5)- 3 = 20 - 3 = 17 Omega`
and the terminal voltage of battery `V = epsi-Ir = 10-0.5xx3 = 10-1.5 = 8.5 V`